数列 1/(n-1)(n+1) 求和
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an=1/(n-1)(n+1)
=1/2((n+1)-(n-1))/((n-1)*(n+1))
=1/2(1/(n-1)-1/(n+1))
即:an=1/2(1/(n-1)-1/(n+1))
a(n-1)=1/2(1/(n-2)-1/(n))
a(n-2)=1/2(1/(n-3)-1/(n-1))
...
a3=1/2(1/2-1/4)
a2=1/2(1-1/3)
a1=1/2(1/0-1/2)未知,本题缺少a1的值
求和就是如上左右两边相加,得
Sn=1/2((1/(n-1)+1/(n-2)+1/(n-3)+...+1/2+1)
-(1/(n+1)+1/(n)+1/(n-1)+...+1/4+1/3)
=1/2(((1/2+1)-(1/(n-1)+1/(n-2))+a1)
=1/2(2/3-(1/(n-1)+1/(n-2))+a1)
=1/2((n+1)-(n-1))/((n-1)*(n+1))
=1/2(1/(n-1)-1/(n+1))
即:an=1/2(1/(n-1)-1/(n+1))
a(n-1)=1/2(1/(n-2)-1/(n))
a(n-2)=1/2(1/(n-3)-1/(n-1))
...
a3=1/2(1/2-1/4)
a2=1/2(1-1/3)
a1=1/2(1/0-1/2)未知,本题缺少a1的值
求和就是如上左右两边相加,得
Sn=1/2((1/(n-1)+1/(n-2)+1/(n-3)+...+1/2+1)
-(1/(n+1)+1/(n)+1/(n-1)+...+1/4+1/3)
=1/2(((1/2+1)-(1/(n-1)+1/(n-2))+a1)
=1/2(2/3-(1/(n-1)+1/(n-2))+a1)
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n取值从2到无穷大
n取值从2到无穷大
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